structure i lecture23
TRANSCRIPT
![Page 1: Structure I Lecture23](https://reader036.vdocuments.mx/reader036/viewer/2022081521/577cce5a1a28ab9e788dd54f/html5/thumbnails/1.jpg)
Structure Analysis II
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Analysis of statically
Indeterminate structures by the
Force MethodPart 1
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Review Conjugate beam
method
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Con
juga
te-B
eam
Sup
port
s
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EIEIM
EIV
BB
BB
5.14062)25(5.562
5.562
'
'
Review Example 1Find the deflection at Point B
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Review Example 2
EIEIEIMCC
162)3(63)1(27'
Find the deflection at Point C
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Determinacy & Stability• Determinacy: when all the forces in
structure can be determined from equilibrium equation , the structure is referred to as statically determinate. Structure having more unknown forces than available equilibrium equations called statically indeterminate
• If n is number of structure parts & r is number of unknown forces:
r = 3n, statically determinate r > 3n, statically indeterminate
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degree 2
ateindetermin Statically )1(3515
nd
nr
degree 4
ateindetermin Statically )2(310210
th
nr
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Force Method: General Procedure
1.Determine the number of redundant reactions
2.Remove enough redundants to form a determinate structure
3.Calculate the displacements that the known loads cause in the determinate structure
4.Calculate displacements at the same points in the determinate structure due to the redundants
5.The sum of the displacements calculated in steps (3) and (4) must be equal to the displacements at the actual indeterminate structure
6.Use equilibrium to determine the remaining reactions
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0
0
BByB
BByBB
BBB
fB
fB
tcoefficienyflexibilitfBB
+
Note 1
Compatibility Equation
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+
0
0
AAAA
AAAAA
AAA
M
M
tcoefficienyflexibilitAA
Note 2
Compatibility Equation
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Note 3
0
0
CCyCByC
BCyBByB
fCfB
fCfB
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Force Method of Analysis: Beams
Example 1Determine the reaction at the roller support B of the beam shown
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Solution
Compatibility Equation
0 BByB fB
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0 BByB fB
EIEIB900010900
Conjugate beam
method
EIEIEI
PLfBB576
3121
3
33
kN 6.15
05769000
0
y
y
BByB
BEI
BEI
fB
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Determine the reaction at the roller support B of the beam shown, the support B settles 1.5 in
Example 2
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125.1 BByB fBCompatibility Equation
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EIEIEIBB
BByB
M
fB3168014401800
125.1
)8()24(
EIEIEIBBB mf 2304144144 )8()24(
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56.5 750
)10(294
3
125.1230431680
125.1
y
EIyEI
BByB
BinI
ksiE
B
fB
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Determine the moment at fixed wall for the beam shown
Example 3
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Solution
EIA3.33
EIAA33.3
k.ft 1000
33.33.33
A
EIAEI
AAAA
MMM
Conjugate beam method